Problem: Evaluate the Maclaurin series. $\dfrac{\pi}{2} -\dfrac{\dfrac{1}{8}{{\pi }^{3}}}{3!}+\dfrac{{\dfrac{1}{32}{\pi }^{5}}}{5!}-...+{{\left( -1 \right)}^{n}}\dfrac{\left(\dfrac{1}{2}\right)^{2n+1}{{\pi }^{2n+1}}}{\left( 2n+1 \right)!}+...$ Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $1$ (Choice C) C $\dfrac13$ (Choice D) D $\dfrac12$
Explanation: Note that the series has terms that alternate in sign, has odd factorials in the denominator, and has odd powers of $~\dfrac{1}{2}\pi~$ in the numerator. That suggests that we can use the Maclaurin series for $~\sin x~$ to find the needed value. Recall that $\sin x=x-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}-...+{{\left( -1 \right)}^{n}}\frac{{{x}^{2n+1}}}{\left( 2n+1 \right)!}+...$ It follows that $\sin\left(\frac{1}{2}x\right)=\frac{x}{2} -\frac{\frac{1}{8}{{x}^{3}}}{3!}+\frac{{\frac{1}{32}{x}^{5}}}{5!}-...+{{\left( -1 \right)}^{n}}\frac{(\frac{1}{2})^{2n+1}{{x }^{2n+1}}}{\left( 2n+1 \right)!}+...$ Hence, the given series converges to $\sin\big( \frac{\pi}{2} \big)$. $\sin \big( \frac{\pi}{2} \big)=1$